I'm trying to implement a simple Lotka-Volterra system in JavaScript, but get different result from what I see in academic papers and slides. This is my equations:
sim2.eval("dxdt(x, y) = (2 * x) - (x * y)");
sim2.eval("dydt(x, y) = (-0.25 * y) + (x * y)");
using coefficients a = 2, b = 1, c = 0.25 and d = 1. Yet, my result looks like this:
when I expected a stable oscillation as seen in these PDF slides:
Could it be the implementation of ndsolve that causes this? Or a machine error in JavaScript due to floating-point arithmetic?
Disregard, the error was simply using a too big evaluation step (dt = 0.1, must be 0.01 at least). The numerical method used is known for this problem.
For serious purposes use a higher order method, the minimum is fixed step classical Runge-Kutta. Then you can also use dt=0.1, it is stable for multiple periods, I tried tfinal=300 without problems. However you will see the step size in the graph as it is visibly piecewise linear. This is much reduced with half the step size, dt=0.05.
function odesolveRK4(f, x0, dt, tmax) {
var n = f.size()[0]; // Number of variables
var x = x0.clone(),xh=[]; // Current values of variables
var dxdt = [], k1=[], k2=[], k3=[], k4=[]; // Temporary variable to hold time-derivatives
var result = []; // Contains entire solution
var nsteps = math.divide(tmax, dt); // Number of time steps
dt2 = math.divide(dt,2);
dt6 = math.divide(dt,6);
for(var i=0; i<nsteps; i++) {
// compute the 4 stages if the classical order-4 Runge-Kutta method
k1 = f.map(function(fj) {return fj.apply(null, x.toArray()); } );
xh = math.add(x, math.multiply(k1, dt2));
k2 = f.map(function(fj) {return fj.apply(null, xh.toArray()); } );
xh = math.add(x, math.multiply(k2, dt2));
k3 = f.map(function(fj) {return fj.apply(null, xh.toArray()); } );
xh = math.add(x, math.multiply(k3, dt));
k4 = f.map(function(fj) {return fj.apply(null, xh.toArray()); } );
x = math.add(x, math.multiply(math.add(math.add(k1,k4), math.multiply(math.add(k2,k3),2)), dt6))
if( 0==i%50) console.log("%3d %o %o",i,dt,x.toString());
result.push(x.clone());
}
return math.matrix(result);
}
math.import({odesolveRK4:odesolveRK4});
Is it possible to get a random number between 1-100 and keep the results mainly within the 40-60 range? I mean, it will go out of that range rarely, but I want it to be mainly within that range... Is it possible with JavaScript/jQuery?
Right now I'm just using the basic Math.random() * 100 + 1.
The simplest way would be to generate two random numbers from 0-50 and add them together.
This gives a distribution biased towards 50, in the same way rolling two dice biases towards 7.
In fact, by using a larger number of "dice" (as #Falco suggests), you can make a closer approximation to a bell-curve:
function weightedRandom(max, numDice) {
let num = 0;
for (let i = 0; i < numDice; i++) {
num += Math.random() * (max/numDice);
}
return num;
}
JSFiddle: http://jsfiddle.net/797qhcza/1/
You have some good answers here that give specific solutions; let me describe for you the general solution. The problem is:
I have a source of more-or-less uniformly distributed random numbers between 0 and 1.
I wish to produce a sequence of random numbers that follow a different distribution.
The general solution to this problem is to work out the quantile function of your desired distribution, and then apply the quantile function to the output of your uniform source.
The quantile function is the inverse of the integral of your desired distribution function. The distribution function is the function where the area under a portion of the curve is equal to the probability that the randomly-chosen item will be in that portion.
I give an example of how to do so here:
http://ericlippert.com/2012/02/21/generating-random-non-uniform-data/
The code in there is in C#, but the principles apply to any language; it should be straightforward to adapt the solution to JavaScript.
Taking arrays of numbers, etc. isn't efficient. You should take a mapping which takes a random number between 0 to 100 and maps to the distribution you need. So in your case, you could take f(x)=-(1/25)x2+4x to get a distribution with the most values in the middle of your range.
I might do something like setup a "chance" for the number to be allowed to go "out of bounds". In this example, a 20% chance the number will be 1-100, otherwise, 40-60:
$(function () {
$('button').click(function () {
var outOfBoundsChance = .2;
var num = 0;
if (Math.random() <= outOfBoundsChance) {
num = getRandomInt(1, 100);
} else {
num = getRandomInt(40, 60);
}
$('#out').text(num);
});
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<button>Generate</button>
<div id="out"></div>
fiddle: http://jsfiddle.net/kbv39s9w/
I needed to solve this problem a few years ago and my solution was easier than any of the other answers.
I generated 3 randoms between the bounds and averaged them. This pulls the result towards the centre but leaves it completely possible to reach the extremities.
It looks stupid but you can use rand twice:
var choice = Math.random() * 3;
var result;
if (choice < 2){
result = Math.random() * 20 + 40; //you have 2/3 chance to go there
}
else {
result = Math.random() * 100 + 1;
}
Sure it is possible. Make a random 1-100. If the number is <30 then generate number in range 1-100 if not generate in range 40-60.
There is a lot of different ways to generate such random numbers. One way to do it is to compute the sum of multiple uniformly random numbers. How many random numbers you sum and what their range is will determine how the final distribution will look.
The more numbers you sum up, the more it will be biased towards the center. Using the sum of 1 random number was already proposed in your question, but as you notice is not biased towards the center of the range. Other answers have propose using the sum of 2 random numbers or the sum of 3 random numbers.
You can get even more bias towards the center of the range by taking the sum of more random numbers. At the extreme you could take the sum of 99 random numbers which each were either 0 or 1. That would be a binomial distribution. (Binomial distributions can in some sense be seen as the discrete version of normal distributions). This can still in theory cover the full range, but it has so much bias towards the center that you should never expect to see it reach the endpoints.
This approach means you can tweak just how much bias you want.
What about using something like this:
var loops = 10;
var tries = 10;
var div = $("#results").html(random());
function random() {
var values = "";
for(var i=0; i < loops; i++) {
var numTries = tries;
do {
var num = Math.floor((Math.random() * 100) + 1);
numTries--;
}
while((num < 40 || num >60) && numTries > 1)
values += num + "<br/>";
}
return values;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
The way I've coded it allows you to set a couple of variables:
loops = number of results
tries = number of times the function will try to get a number between 40-60 before it stops running through the while loop
Added bonus: It uses do while!!! Awesomeness at its best
You can write a function that maps random values between [0, 1) to [1, 100] according to weight. Consider this example:
Here, the value 0.95 maps to value between [61, 100].
In fact we have .05 / .1 = 0.5, which, when mapped to [61, 100], yields 81.
Here is the function:
/*
* Function that returns a function that maps random number to value according to map of probability
*/
function createDistributionFunction(data) {
// cache data + some pre-calculations
var cache = [];
var i;
for (i = 0; i < data.length; i++) {
cache[i] = {};
cache[i].valueMin = data[i].values[0];
cache[i].valueMax = data[i].values[1];
cache[i].rangeMin = i === 0 ? 0 : cache[i - 1].rangeMax;
cache[i].rangeMax = cache[i].rangeMin + data[i].weight;
}
return function(random) {
var value;
for (i = 0; i < cache.length; i++) {
// this maps random number to the bracket and the value inside that bracket
if (cache[i].rangeMin <= random && random < cache[i].rangeMax) {
value = (random - cache[i].rangeMin) / (cache[i].rangeMax - cache[i].rangeMin);
value *= cache[i].valueMax - cache[i].valueMin + 1;
value += cache[i].valueMin;
return Math.floor(value);
}
}
};
}
/*
* Example usage
*/
var distributionFunction = createDistributionFunction([
{ weight: 0.1, values: [1, 40] },
{ weight: 0.8, values: [41, 60] },
{ weight: 0.1, values: [61, 100] }
]);
/*
* Test the example and draw results using Google charts API
*/
function testAndDrawResult() {
var counts = [];
var i;
var value;
// run the function in a loop and count the number of occurrences of each value
for (i = 0; i < 10000; i++) {
value = distributionFunction(Math.random());
counts[value] = (counts[value] || 0) + 1;
}
// convert results to datatable and display
var data = new google.visualization.DataTable();
data.addColumn("number", "Value");
data.addColumn("number", "Count");
for (value = 0; value < counts.length; value++) {
if (counts[value] !== undefined) {
data.addRow([value, counts[value]]);
}
}
var chart = new google.visualization.ColumnChart(document.getElementById("chart"));
chart.draw(data);
}
google.load("visualization", "1", { packages: ["corechart"] });
google.setOnLoadCallback(testAndDrawResult);
<script src="https://www.google.com/jsapi"></script>
<div id="chart"></div>
Here's a weighted solution at 3/4 40-60 and 1/4 outside that range.
function weighted() {
var w = 4;
// number 1 to w
var r = Math.floor(Math.random() * w) + 1;
if (r === 1) { // 1/w goes to outside 40-60
var n = Math.floor(Math.random() * 80) + 1;
if (n >= 40 && n <= 60) n += 40;
return n
}
// w-1/w goes to 40-60 range.
return Math.floor(Math.random() * 21) + 40;
}
function test() {
var counts = [];
for (var i = 0; i < 2000; i++) {
var n = weighted();
if (!counts[n]) counts[n] = 0;
counts[n] ++;
}
var output = document.getElementById('output');
var o = "";
for (var i = 1; i <= 100; i++) {
o += i + " - " + (counts[i] | 0) + "\n";
}
output.innerHTML = o;
}
test();
<pre id="output"></pre>
Ok, so I decided to add another answer because I felt like my last answer, as well as most answers here, use some sort of half-statistical way of obtaining a bell-curve type result return. The code I provide below works the same way as when you roll a dice. Therefore, it is hardest to get 1 or 99, but easiest to get 50.
var loops = 10; //Number of numbers generated
var min = 1,
max = 50;
var div = $("#results").html(random());
function random() {
var values = "";
for (var i = 0; i < loops; i++) {
var one = generate();
var two = generate();
var ans = one + two - 1;
var num = values += ans + "<br/>";
}
return values;
}
function generate() {
return Math.floor((Math.random() * (max - min + 1)) + min);
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
I'd recommend using the beta distribution to generate a number between 0-1, then scale it up. It's quite flexible and can create many different shapes of distributions.
Here's a quick and dirty sampler:
rbeta = function(alpha, beta) {
var a = 0
for(var i = 0; i < alpha; i++)
a -= Math.log(Math.random())
var b = 0
for(var i = 0; i < beta; i++)
b -= Math.log(Math.random())
return Math.ceil(100 * a / (a+b))
}
var randNum;
// generate random number from 1-5
var freq = Math.floor(Math.random() * (6 - 1) + 1);
// focus on 40-60 if the number is odd (1,3, or 5)
// this should happen %60 of the time
if (freq % 2){
randNum = Math.floor(Math.random() * (60 - 40) + 40);
}
else {
randNum = Math.floor(Math.random() * (100 - 1) + 1);
}
The best solution targeting this very problem is the one proposed by BlueRaja - Danny Pflughoeft but I think a somewhat faster and more general solution is also worth mentioning.
When I have to generate random numbers (strings, coordinate pairs, etc.) satisfying the two requirements of
The result set is quite small. (not larger than 16K numbers)
The result set is discreet. (like integer numbers only)
I usually start by creating an array of numbers (strings, coordinate pairs, etc.) fulfilling the requirement (In your case: an array of numbers containing the more probable ones multiple times.), then choose a random item of that array. This way, you only have to call the expensive random function once per item.
Distribution
5% for [ 0,39]
90% for [40,59]
5% for [60,99]
Solution
var f = Math.random();
if (f < 0.05) return random(0,39);
else if (f < 0.95) return random(40,59);
else return random(60,99);
Generic Solution
random_choose([series(0,39),series(40,59),series(60,99)],[0.05,0.90,0.05]);
function random_choose (collections,probabilities)
{
var acc = 0.00;
var r1 = Math.random();
var r2 = Math.random();
for (var i = 0; i < probabilities.length; i++)
{
acc += probabilities[i];
if (r1 < acc)
return collections[i][Math.floor(r2*collections[i].length)];
}
return (-1);
}
function series(min,max)
{
var i = min; var s = [];
while (s[s.length-1] < max) s[s.length]=i++;
return s;
}
You can use a helper random number to whether generate random numbers in 40-60 or 1-100:
// 90% of random numbers should be between 40 to 60.
var weight_percentage = 90;
var focuse_on_center = ( (Math.random() * 100) < weight_percentage );
if(focuse_on_center)
{
// generate a random number within the 40-60 range.
alert (40 + Math.random() * 20 + 1);
}
else
{
// generate a random number within the 1-100 range.
alert (Math.random() * 100 + 1);
}
If you can use the gaussian function, use it. This function returns normal number with average 0 and sigma 1.
95% of this number are within average +/- 2*sigma. Your average = 50, and sigma = 5 so
randomNumber = 50 + 5*gaussian()
The best way to do that is generating a random number that is distributed equally in a certain set of numbers, and then apply a projection function to the set between 0 and a 100 where the projection is more likely to hit the numbers you want.
Typically the mathematical way of achieving this is plotting a probability function of the numbers you want. We could use the bell curve, but let's for the sake of easier calculation just work with a flipped parabola.
Let's make a parabola such that its roots are at 0 and 100 without skewing it. We get the following equation:
f(x) = -(x-0)(x-100) = -x * (x-100) = -x^2 + 100x
Now, all the area under the curve between 0 and 100 is representative of our first set where we want the numbers generated. There, the generation is completely random. So, all we need to do is find the bounds of our first set.
The lower bound is, of course, 0. The upper bound is the integral of our function at 100, which is
F(x) = -x^3/3 + 50x^2
F(100) = 500,000/3 = 166,666.66666 (let's just use 166,666, because rounding up would make the target out of bounds)
So we know that we need to generate a number somewhere between 0 and 166,666. Then, we simply need to take that number and project it to our second set, which is between 0 and 100.
We know that the random number we generated is some integral of our parabola with an input x between 0 and 100. That means that we simply have to assume that the random number is the result of F(x), and solve for x.
In this case, F(x) is a cubic equation, and in the form F(x) = ax^3 + bx^2 + cx + d = 0, the following statements are true:
a = -1/3
b = 50
c = 0
d = -1 * (your random number)
Solving this for x yields you the actual random number your are looking for, which is guaranteed to be in the [0, 100] range and a much higher likelihood to be close to the center than the edges.
This answer is really good. But I would like to post implementation instructions (I'm not into JavaScript, so I hope you will understand) for different situation.
Assume you have ranges and weights for every range:
ranges - [1, 20], [21, 40], [41, 60], [61, 100]
weights - {1, 2, 100, 5}
Initial Static Information, could be cached:
Sum of all weights (108 in sample)
Range selection boundaries. It basically this formula: Boundary[n] = Boundary[n - 1] + weigh[n - 1] and Boundary[0] = 0. Sample has Boundary = {0, 1, 3, 103, 108}
Number generation:
Generate random number N from range [0, Sum of all weights).
for (i = 0; i < size(Boundary) && N > Boundary[i + 1]; ++i)
Take ith range and generate random number in that range.
Additional note for performance optimizations. Ranges don't have to be ordered neither ascending nor descending order, so for faster range look-up range that has highest weight should go first and one with lowest weight should go last.
I have some code which takes strings representing hexadecimal numbers - hex colors, actually - and adds them. For example, adding aaaaaa and 010101 gives the output ababab.
However, my method seems unnecessarily long and complicated:
var hexValue = "aaaaaa";
hexValue = "0x" + hexValue;
hexValue = parseInt(hexValue, 16);
hexValue = hexValue + 0x010101;
hexValue = hexValue.toString(16);
document.write(hexValue); // outputs 'ababab'
The hex value is still a string after concatenating 0x, so then I have to change it to a number, then I can add, and then I have to change it back into hex format! There are even more steps if the number I'm adding to it is a hexadecimal string to begin with, or if you take into consideration that I am removing the # from the hex color before all this starts.
Surely there's a simpler way to do such simple hexadecimal calculations! And just to be clear, I don't mean just putting it all on one line like (parseInt("0x"+"aaaaaa",16)+0x010101).toString(16) or using shorthand - I mean actually doing less operations.
Is there some way to get javascript to stop using decimal for all of its mathematical operations and use hex instead? Or is there some other method of making JS work with hex more easily?
No, there is no way to tell the JavaScript language to use hex integer format instead of decimal by default. Your code is about as concise as it gets but note that you do not need to prepend the "0x" base indicator when you use "parseInt" with a base.
Here is how I would approach your problem:
function addHexColor(c1, c2) {
var hexStr = (parseInt(c1, 16) + parseInt(c2, 16)).toString(16);
while (hexStr.length < 6) { hexStr = '0' + hexStr; } // Zero pad.
return hexStr;
}
addHexColor('aaaaaa', '010101'); // => 'ababab'
addHexColor('010101', '010101'); // => '020202'
As mentioned by a commenter, the above solution is chock full of problems, so below is a function that does proper input validation and adds color channels separately while checking for overflow.
function addHexColor2(c1, c2) {
const octetsRegex = /^([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})$/i
const m1 = c1.match(octetsRegex)
const m2 = c2.match(octetsRegex)
if (!m1 || !m2) {
throw new Error(`invalid hex color triplet(s): ${c1} / ${c2}`)
}
return [1, 2, 3].map(i => {
const sum = parseInt(m1[i], 16) + parseInt(m2[i], 16)
if (sum > 0xff) {
throw new Error(`octet ${i} overflow: ${m1[i]}+${m2[i]}=${sum.toString(16)}`)
}
return sum.toString(16).padStart(2, '0')
}).join('')
}
addHexColor2('aaaaaa', 'bogus!') // => Error: invalid hex color triplet(s): aaaaaa / bogus!
addHexColor2('aaaaaa', '606060') // => Error: octet 1 overflow: aa+60=10a
How about this:
var hexValue = "aaaaaa";
hexValue = (parseInt(hexValue, 16) + 0x010101).toString(16);
document.writeln(hexValue); // outputs 'ababab'
There is no need to add the 0x prefix if you use parseInt.
I think accepted answer is wrong. Hexadecimal color representation is not a linear. But instead, 3 sets of two characters are given to R, G & B.
So you can't just add a whole number and expect to RGB to add up correctly.
For Example
n1 = '005500'; <--- green
n2 = '00ff00'; <--- brighter green
Adding these numbers should result in a greener green.
In no way, adding greens should increase RED to increase. but by doing what accepted answer is doing, as in just treat whole number as one number then you'd carry over for numbers adding upto greater than f, f+1 = 10.
you get `015400` so by adding greens the RED increased .... WRONG
adding 005500 + 00ff00 should result in, = 00ff00. You can't add more green to max green.
For folks looking for a function that can add and subtract HEX colors without going out of bounds on an individual tuple, I wrote this function a few minutes ago to do just that:
export function shiftColor(base, change, direction) {
const colorRegEx = /^\#?[A-Fa-f0-9]{6}$/;
// Missing parameter(s)
if (!base || !change) {
return '000000';
}
// Invalid parameter(s)
if (!base.match(colorRegEx) || !change.match(colorRegEx)) {
return '000000';
}
// Remove any '#'s
base = base.replace(/\#/g, '');
change = change.replace(/\#/g, '');
// Build new color
let newColor = '';
for (let i = 0; i < 3; i++) {
const basePiece = parseInt(base.substring(i * 2, i * 2 + 2), 16);
const changePiece = parseInt(change.substring(i * 2, i * 2 + 2), 16);
let newPiece = '';
if (direction === 'add') {
newPiece = (basePiece + changePiece);
newPiece = newPiece > 255 ? 255 : newPiece;
}
if (direction === 'sub') {
newPiece = (basePiece - changePiece);
newPiece = newPiece < 0 ? 0 : newPiece;
}
newPiece = newPiece.toString(16);
newPiece = newPiece.length < 2 ? '0' + newPiece : newPiece;
newColor += newPiece;
}
return newColor;
}
You pass your base color as parameter 1, your change as parameter 2, and then 'add' or 'sub' as the last parameter depending on your intent.
Can I write a JavaScript function from scratch that behaves like Math.random?
(By that I mean without using Math.random.)
Yes you can, you can implement your own LCG number generator, but as Sarnold mentions you need to maintain state between calls.
Per #OscarGomez's answer regarding a linear congruential generator, here's an example of a random number generator as a plain JavaScript function. Of course, its quality of "randomness" (currently very poor due to a short cycle) appears to be dependent on picking good values for the constants in the enclosed object "o".
var random = (function() {
var o = {mod: 13, mul: 11, inc: 7, x: 0};
return function() {
return o.x = (o.mul * o.x + o.inc) % o.mod
}
})();
random(); // => 7
random(); // => 6
random(); // => 8
random(); // => 4
Here's a more portable version which can have separate generator instances and seeds:
function Random(s) {
this.seed = s || 0;
this.mod = 13;
this.mul = 11;
this.inc = 7;
this.x = this.seed;
}
Random.prototype.next = function() {
return (this.x = (this.mul * this.x + this.inc) % this.mod);
};
var r = new Random(1);
r.next(); // => 5
r.next(); // => 10
r.next(); // => 7
What I usually do when I need some randomness but am to lazy to look up the syntax is to implement the logistic map (a discrete chaotic system). In pseudo code this is like this:
var x = 0.234; // or some other number between 0 and 1 ( but not 0.5 )
for (var n=1; n<=100;n++){
x = 4 * x * (1-x); // this is the iteration
console.log(x);
}
This would print 100 somehow random numbers, not really random but for many situations random enough.
Sorry for not giving a javascript answer, I havent used that for a decade.
I'm looking for the best way of implementing random number generator, that will allow me to have control over probability from what range the generated number will be returned. To visualize what I'm trying to achieve I have a picture :
So to summarize :
Let's say that my range is 400. At the beginning I'd like to have 5% probability of getting number 0-20. But at some moment in time I'd like to have this probability increased up to 50%. Hope you get the idea.
Hmm, working on your original I had a pretty simple algorithm to generate ranges in an array in the appropriate proportion, then randomly select a range and generate a random number within that range. No doubt it can be optimised if necessary, but it works for me.
It looks like a lot of code, but 3/4 of it is comments, test data and function, the actual randomRange function is only 17 lines of code.
<script type="text/javascript">
function randomRange(dataArray) {
// Helper function
function getRandomInRange(s, f) {
return (Math.random() * (f-s+1) | 0) + s
}
// Generate new data array based on probability
var i, j = dataArray.length;
var oArray = [];
var o;
while (j--) {
o = dataArray[j];
// Make sure probability is an integer
for (i=0, iLen=o.probability|0; i<iLen; i++) {
oArray.push([o.rangeStart, o.rangeEnd]);
}
}
// Randomly select a range from new data array and
// generate a random number in that range
var oEnd = oArray.length;
var range = oArray[getRandomInRange(0, oArray.length - 1)];
return getRandomInRange(range[0], range[1]);
}
// Test data set. Probability just has to be
// representative, so 50/50 === 1/1
var dataArray = [
{
rangeStart: 0,
rangeEnd : 20,
probability: 1
},
{
rangeStart: 21,
rangeEnd : 400,
probability: 1
}
];
// Test function to show range and number is randomly
// selected for given probability
function testIt() {
var el0 = document.getElementById('div0');
var el1 = document.getElementById('div1');
function run() {
var n = randomRange(dataArray);
if (n <= 20) {
el0.innerHTML += '*';
} else {
el1.innerHTML += '*';
}
}
setInterval(run, 500);
}
</script>
<button onclick="testIt();">Generate random number</button>
<div>Numbers 0 - 20</div>
<div id="div0"></div>
<div>Numbers 21 - 400</div>
<div id="div1"></div>
It sounds to me like what you're looking for is a way to generate numbers on a normal (or Gaussian) distribution (take a look at the Wikipedia page if you don't know what that means).
The Box-Muller transformation can be used to generate pairs of normally distributed numbers.
Here is a c++ implementation of the polar form of the Box-Muller transformation that shouldn't be hard to translate to javascript.
// Return a real number from a normal (Gaussian) distribution with given
// mean and standard deviation by polar form of Box-Muller transformation
double x, y, r;
do
{
x = 2.0 * rand() - 1.0;
y = 2.0 * rand() - 1.0;
r = x * x + y * y;
}
while ( r >= 1.0 || r == 0.0 );
double s = sqrt( -2.0 * log(r) / r );
return mean + x * s * stddev;
Where mean is the mean of the normal distribution and stddev is the Standard Deviation of the distribution. This code is from a MersesenneTwister C++ class that I've been using recently that you can find on Rick Wagner's page. You can find some more useful information about the Box-Muller transformation on this page.